There seem to be at least three solutions -- two with a 2 in r8c7 (in this case the six cells at r3c5, r3c6, r4c1, r4c5, r5c1, r5c6 can be filled in either of two ways), and one with a 1 in r8c7 and a 2 in r9c7. There may be additional solutions with a 1 in r8c7 and a 1 in r9c7.

Thank you for your hint. I must had made a mistake in painting the picture. I can remember the methods used to solve it but I can't find my original grid. I will give another sudoku of this variants some days later.

You're welcome. From this point on, though, I had to use trial and error (something I'm never opposed to when solving tough puzzles). If you find something better, please let me know.

Since there must be a 3 in either r5c7 or r5c9, there cannot be a 3 in either r6c7 or r6c9, which means there must be a 3 in either r6c4 or r6c5. One of these leads (eventually) to a contradiction, the other to a solution.

If you take a valid, completed, regular Sudoku grid, and change all the 4's and 7's to 1's, and all the 5's and 8's to 2's, and all the 6's and 9's to 3's, you have a valid, completed, 123 Sudoku grid.

But the original clue set, converted as above, will probably no longer be sufficient to establish a unique solution.

As I recall, the maximum number of clues in an independent clue set is known to be somewhere in the 30's, and the thought-to-be-minimum is 17. Has anybody given any thought (shudder) to what the maximum and minimum might be in the 123 case? I assume both would be higher than for regular Sudoku.

There is a tacit agreement that there is a difference between 1-2-3 sudoku, 2-3-4 sudoku, 1-2-3-4 sudoku, 1-2-3-4-5 sudoku in one hand and 123 sudoku, 1234 sudoku in the other . The first group is where we have one 1, two 2's, three 3's or four 4's and the other group has the same number of 1's, 2's, 3's or 4's.

For proper closure of this thread I'll show all the possible solutions for this (now deleted) 123c puzzle... Turns out you could work out 58 cells and then there are altogether 5 different solutions coming out from the remaining 23 cells...

Original puzzle:

All the 58 cells you could work out (and candidates for the remaining 23):